Q.Evaluate the definite integral: $\int_{\pi/6}^{\pi/4} \operatorname{cosec} x \ dx$

Idea: use the standard antiderivative $\int\operatorname{cosec}x\,dx=\log\big|\tan\tfrac{x}{2}\big|+C$ and plug in the limits.

Standard values: $\tan\tfrac{\pi}{8}=\sqrt2-1$ and $\tan\tfrac{\pi}{12}=2-\sqrt3$. Therefore …

The antiderivative

The standard result is

so

Evaluate at the upper limit $x=\pi/4$

Use the half-angle identity $\tan\tfrac{\theta}{2}=\dfrac{1-\cos\theta}{\sin\theta}$ with $\theta=\tfrac{\pi}{4}$ ($\cos\tfrac{\pi}{4}=\sin\tfrac{\pi}{4}=\tfrac{\sqrt2}{2}$):

So the upper value is $\log(\sqrt2-1)$.

Evaluate at the lower limit $x=\pi/6$

With $\theta=\tfrac{\pi}{6}$ ($\cos\tfrac{\pi}{6}=\tfrac{\sqrt3}{2}$, $\sin\tfrac{\pi}{6}=\tfrac12$):

So the lower value is $\log(2-\sqrt3)$. …

Method: Definite integral of $\operatorname{cosec} x$ via its logarithmic antiderivative

Use $\int\operatorname{cosec} x\,dx=\log\big|\tan\dfrac{x}{2}\big|$ (equivalently $-\log|\operatorname{cosec} x+\cot x|$), then evaluate.

Steps

Step 1: Recall a standard antiderivative. $\int\operatorname{cosec} x\,dx=\log\big|\tan\tfrac{x}{2}\big|+C$.

Step 2: Bracket across the limits.

$\int_{a}^{b}\operatorname{cosec} x\,dx=\Big[\log\big|\tan\tfrac{x}{2}\big|\Big]_{a}^{b}.$ …

Common Mistakes

Mistake 1: Misremembering $\int\operatorname{cosec} x\,dx$ as $\log|\operatorname{cosec} x+\cot x|$ (wrong sign).

Why it's wrong: the correct form is $-\log|\operatorname{cosec} x+\cot x|=\log|\tan\tfrac{x}{2}|$. Correct approach: keep the minus, or use the half-angle form.

Mistake 2: Turning $\log\tan\tfrac{b}{2}-\log\tan\tfrac{a}{2}$ into a quotient of logs. …